Question: The equation of hyperbola $H$ is $\dfrac {(y+2)^{2}}{16}-\dfrac {(x+2)^{2}}{4} = 1$. What are the asymptotes?
Explanation: We want to rewrite the equation in terms of $y$ , so start off by moving the $y$ terms to one side: $\dfrac {(y+2)^{2}}{16} = 1 + \dfrac {(x+2)^{2}}{4}$ Multiply both sides of the equation by $16$ $(y+2)^{2} = { 16 + \dfrac{ (x+2)^{2} \cdot 16 }{4}}$ Take the square root of both sides. $\sqrt{(y+2)^{2}} = \pm \sqrt { 16 + \dfrac{ (x+2)^{2} \cdot 16 }{4}}$ $ y + 2 = \pm \sqrt { 16 + \dfrac{ (x+2)^{2} \cdot 16 }{4}}$ As $x$ approaches positive or negative infinity, the constant term in the square root matters less and less, so we can just ignore it. $y + 2 \approx \pm \sqrt {\dfrac{ (x+2)^{2} \cdot 16 }{4}}$ $y + 2 \approx \pm \left(\dfrac{4 \cdot (x + 2)}{2}\right)$ Subtract $2$ from both sides and rewrite as an equality in terms of $y$ to get the equation of the asymptotes: $y = \pm \dfrac{2}{1}(x + 2) -2$